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Capacitors

JEE Main level, numerical answer, medium

158 of 180

Question 158 of 180

Two parallel plate capacitors ${C}_{1}$ and ${C}_{2}$ each having capacitance of $10\mu F$ are individually charged by a $100V$ D.C. source. Capacitor ${C}_{1}$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor ${C}_{2}$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor ${C}_{1}$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be _____ $V$ .
(Assuming Dielectric constant $=10$ )

Show answer and solution

Answer: 55

As source is connected in the process therefore, additional charge will be provided. Therefore, charge on

{C}_{1}=K\left(CV\right)

.
Since source is disconnected, so its charge will remain the same. Then, charge on

{C}_{2}=CV

.
New capacitance in both cases will become,

KC

.
When they are connected in parallel charge will be equally divided so charge on one capacitor is

q=\left(\dfrac{K+1}{2}\right)CV

.
So common potential,

{V}^{'}=\dfrac{q}{KC}=\dfrac{K+1}{2K}V=55V

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JEE Main 2025 Apr 2 Shift 1 question paper

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